Re: Return value from a binding operator?

Hi Chris,

Think of the middle match as a substition.

the match basically looks to match all of the string after the last / character. The end of line anchor $ is included in the capturing group, where we may be used to seeing this outside of the capturing group.

It does not so much as as strip the path off, rather matches the final part of the path, or the basename. I don't know if including the eol anchor $ retrieves the eol character or makes no difference.

 m!        #start match
  /?       # 0 or 1 '/'
  (         # start capture $1
   [^/]     #consisting of no '/' characters
    +         # 1 or more times 
   $         # until and including eol (?)
  )         # end capture $1
 !         # end match
[download]

this would also match a path with no directories.

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Re^2: Return value from a binding operator? by Spinone (Novice) on Apr 12, 2013 at 23:07 UTC
Really useful, thanks. I'd worked out that /? meant "optional /" (though not why that was relevant) but, as an absolute Perl beginner, I had assumed that the caret slash (^/) meant "a line beginning with a slash" (still not sure why it doesn't), which didn't make any sense, so I obviously need to get to grips with that. Lots to learn.... Chris	[reply]
Re^3: Return value from a binding operator? by dsheroh (Monsignor) on Apr 13, 2013 at 09:50 UTC
The `^` metacharacter has different meanings in different parts of a regex. Normally, it does mean "start of line", as you thought. If the regex were `m!^/!`, then it would match lines beginning with `/`. However, as the first character in a character class, it negates the class, so `[^/]` means "match any character except `/`". (But, again, that's only if it's the first character in the class - `[/^]` means "match either a `/` or a `^`". And `[^/^]` would match any character except those two.)	[reply] [d/l] [select]


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